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Chapter 9
Poincaré Models of Hyperbolic
Geometry
9.1
The Poincaré Upper Half Plane Model
The next model of the hyperbolic plane that we will consider is also due to Henri Poincaré.
We will be using the upper half plane, or {(x, y) | y > 0}. We will want to think of this
with a different distance metric on it.
Let H = {x + iy | y > 0} together with the arclength element
p
dx2 + dy 2
ds =
.
y
Note that we have changed the arclength element for this model!!!
9.2
Vertical Lines
Let x(t) = (x(t), y(t)) be a piecewise smooth parameterization of a curve between the points
x(t0 ) and x(t1 ).
Recall that in order to find the length of a curve we break the curve into small pieces
and approximate the curve by multiple line
p segments. In the limiting process we find that
the Euclidean arclength element is ds = dx2 + dy 2 . We then find the length of a curve
by integrating the arclength over the parameterization of the curve.
s
Z t1 µ ¶2 µ ¶2
dx
dy
s=
+
dt.
dt
dt
t0
Now, we want to work in the Poincaré Half Plane model. In this case the length of this
same curve would be
r
¡ dx ¢2 ³ dy ´2
Z t1
+ dt
dt
sP =
dt.
y
t0
Let’s look at this for a vertical line segment from (x0 , y0 ) to (x0 , y1 ). We need to parameterize the curve, and then use the arclength element to find its length. Its parameterization
is:
x(t) = (x0 , y), y ∈ [y0 , y1 ].
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9.3. ISOMETRIES
67
The Poincaré arclength is then
r
¡ dx ¢2 ³ dy ´2
Z t1
Z t1
+ dt
dt
1
sP =
dt =
dy = ln(y)|yy10 = ln(y1 ) − ln(y0 ) = ln(y1 /y0 )
y
y
t0
t0
Now, consider any piecewise smooth curve x(t) = (x(t), y(t)) starting at (x0 , y0 ) and
ending at (x0 , y1 ). So this curves starts and ends at the same points as this vertical line
segment. Suppose that y(t) is an increasing function. This is reasonable. Now, we have
r
¡ dx ¢2 ³ dy ´2
Z t1
+ dt
dt
s=
dt
y
t0
r³ ´
2
dy
Z t1
dt
dt
≥
y
t0
Z y(t1 )
dy
≥
y(t0 ) y
≥ ln(y(t1 )) − ln(y(t0 )).
This means that this curve is longer than the vertical line segment which joins the two
points. Therefore, the shortest path that joins these two points is a vertical (Euclidean)
line segment. Thus, vertical (Euclidean) lines in the upper half plane are lines in the
Poincaré model.
Let’s find the distance from (1, 1) to (1, 0) which would be the distance to the real axis.
Now, since (1, 0) is NOT a point of H , we need to find lim d((1, 1), (1, δ)). According to
δ→0
what we have above,
dP ((1, 1), (1, δ)) = ln(1) − ln(δ) = − ln(δ).
Now, in the limit we find that
dP ((1, 1), (1, 0)) = lim dP ((1, 1), (1, δ)) = lim − ln(δ) = +∞
δ→0
δ→0
This tells us that a vertical line has infinite extent in either direction.
9.3
Isometries
Recall that an isometry is a map that preserves distance. What are the isometries of H ?
The arclength element must be preserved under the action of any isometry. That is, a
map
(u(x, y), v(x, y))
is an isometry if
du2 + dv 2
dx2 + dy 2
=
.
v2
y2
Some maps will be obvious candidates for isometries and some will not.
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Let’s start with the following candidate:
Ta (x, y) = (u, v) = (x + a, y).
Now, clearly du = dx and dv = dy, so
du2 + dv 2
dx2 + dy 2
=
.
v2
y2
Thus, Ta is an isometry. What does it do? It translates the point a units in the horizontal
direction. This is called the horizontal translation by a.
Let’s try:
Rb (x, y) = (u, v) = (2b − x, y).
Again, du = −dx, dv = dy and our arclength element is preserved. This isometry is a
reflection through the vertical line x = b.
We need to consider the following map:
¶
µ
y
x
,
.
Φ(x, y) = (u, v) =
x2 + y 2 x2 + y 2
First, let’s check that it is a Poincaré isometry. Let r2 = x2 + y 2 . Then
õ
¶2 µ 2
¶2 !
du2 + dv 2
r2 dx − 2x2 dx − 2xydy
r4
r dy − 2xydx − 2y 2 dy
= 2
+
v2
y
r4
r4
µ 2
¶
1 ((y − x2 )dx − 2xydy)2 − ((x2 − y 2 )dy − 2xydx)2
= 2
y
r4
¢
1 ¡
= 4 2 (x4 − 2x2 y 2 + y 4 + 4x2 y 2 )dx2 − (2xy(y 2 − x2 ) + 2xy(x2 − y 2 )dxdy + r4 dy 2
r y
dx2 + dy 2
=
y2
We will study this function further. It is called inversion in the unit circle.
9.4
Inversion in the Circle: Euclidean Considerations
We are building a tool that we will use in studying H . This is a Euclidean tool, so we will
be working in Euclidean geometry to prove results about this tool.
Let’s look at this last isometry. Note what this function does. For each point (x, y),
let r2 = x2 + y 2 . This makes r the distance from the origin to (x, y). This function sends
(x, y) to (x/r2 , y/r2 ). The distance from Φ(x, y) = (x/r2 , y/r2 ) to the origin is 1/r2 . Thus,
if r > 1 then the image of the point is on the same ray, but its distance to the origin is now
less than one. Likewise, if r < 1, then the image lies on the same ray but the image point
lies at a distance greater than 1 from the origin. If r = 1, then Φ(x, y) = (x, y). Thus, Φ
leaves the unit circle fixed and sends every point inside the unit circle outside the circle and
every point outside the unit circle gets sent inside the unit circle. In other words, Φ turns
the circle inside out.
What does Φ do to a line? What does it do to a circle? Let’s see.
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9.4. INVERSION IN THE CIRCLE: EUCLIDEAN CONSIDERATIONS
69
The image of a point P under inversion in a circle centered at O and with radius r is
the point P 0 on the ray OP and such that
|OP 0 | =
r2
.
|OP |
Lemma 9.1 Let ` be a line which does not go through the origin O. The image of ` under
inversion in the unit circle is a circle which goes through the origin O.
Proof: We will prove this for a line ` not intersecting the unit circle.
Let A be the foot of O on ` and
let |OA| = a. Find A0 on OA so
that |OA0 | = 1/a. Construct the circle with diameter OA0 . We want to
show that this circle is the image of
` under inversion.
Let P ∈ ` and let |OP | = p.
O
Let P 0 be the intersection of the segP'
ment OP with the circle with diameter OA0 . Let |OP 0 | = x. Now,
look at the two triangles 4OAP and
A'
4OP 0 A0 . These two Euclidean triangles are similar, so
P
A
|OA|
|OP 0 |
=
|OA0 |
|OP |
x
a
=
1/a
p
1
x=
p
Therefore, P 0 is the image of P under inversion in the unit circle.
Lemma 9.2 Suppose Γ is a circle
which does not go through the origin O. Then the image of Γ under inversion in the unit
circle is a circle.
Proof: Again, I will prove this for just one case: the case where Γ does not intersect the
unit circle.
Let the line through O and the center of Γ intersect Γ at points A and B. Let |OA| = a
and |OB| = b. Let Γ0 be the image of Γ under dilation by the factor 1/ab. This dilation is
∆ : (x, y) 7→ (x/ab, y/ab).
Let B 0 and A0 be the images of A and B, respectively, under this dilation, i.e. ∆(A) = B 0
and ∆(B) = A0 . Then |OA0 | = (1/ab)b = 1/a and |OB 0 | = (1/ab)a = 1/b. Thus, A0 is the
image of A under inversion in the unit circle. Likewise, B 0 is the image of B. Let `0 be an
arbitrary ra through O which intersects Γ at P and Q. Let Q0 and P 0 be the images of P
and Q, respectively, under the dilation, ∆.
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P
Q
Q'
P'
O
B'
A'
A
B
Now, 4OA0 P 0 ∼ 4OBQ, since
one is the dilation of the other. Note
that ∠QBA ∼
= ∠QP A by the Star
Trek lemma, and hence 4OBQ ∼
4OP A. Thus, 4OA0 P 0 ∼ 4OP A.
From this it follows that
Γ′
Γ
Figure 9.1:
|OA0 |
|OP 0 |
=
|OP |
|OA|
1/a
|OP 0 |
=
|OP |
a
1
|OP 0 | =
|OP |
Thus, P 0 is the image of P under inversion, and Γ0 is the image of Γ under inversion.
Lemma 9.3 Inversions preserve angles.
Proof: We will just consider the
case of an angle α created by the in{ α
tersection of a line ` not intersecting
0
A {'
the unit circle, and a line ` through
O.
{''
Let A be the vertex of the angle
A'
β
α. Let P be the foot of O in `. Let
O
P 0 be the image of P under inversion. Then the image of ` is a circle
P'
Γ whose diameter isTOP 0 . The image of A is A0 = Γ `0 . Let `00 be
P
the tangent to Γ at A0 . Then β, the
angle formed by `0 and `00 at A0 is
the image of α under inversion. We
need to show that α ∼
= β.
First, 4OAP ∼ 4OP 0 A0 , since
Figure 9.2:
they are both right triangles and
share the angle O. Thus, ∠A0 P 0 O ∼
= ∠OAP ∼
= α. By the tangential case of the Star
0
0
∼
∼
Trek lemma, β = ∠A P O. Thus, α = β.
9.5
Lines in the Poincaré Half Plane
From what we have just shown we can now prove the following.
Lemma 9.4 Lines in the Poincaré upper half plane model are (Euclidean) lines and (Euclidean) half circles that are perpendicular to the x-axis.
Proof: Let P and Q be points in H not on the same vertical line. Let Γ be the circle
through P and Q whose center lies on the x-axis. Let Γ intersect the x-axis at M and
N . Now consider the mapping ϕ which is the composition of a horizontal translation by
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9.5. LINES IN THE POINCARÉ HALF PLANE
71
−M followed by inversion in the unit circle. This map ϕ is an isometry because it is the
composition of two isometries. Note that M is first sent to O and then to ∞ by inversion.
Thus, the image of Γ is a (Euclidean) line. Since the center of the circle is on the real axis,
the circle intersects the axis at right angles. Since inversion preserves angles, the image of
Γ is a vertical (Euclidean) line. Since vertical lines are lines in the model, and isometries
preserve arclength, it follows that Γ is a line through P and Q.
Problem: Let P = 4 + 4i and Q = 5 + 3i. We want to find M , N , and the distance from
P to Q.
First we need to find Γ. We need to find the perpendicular bisector of the segment P Q
and then find where this intersects the real axis. The midpoint of P Q is the point (9+7i)/2,
or (9/2, 7/2). The equation of the line through P Q is y = 8 − x. Thus, the equation of the
perpendicular bisector is y = x − 1. This intersects the x-axis at x = 1, so the center of the
circle is 1 + 0i. The circle has to go through the points 4 + 4i and 5 + 3i. Thus the radius is
5, using the Pythagorean theorem. Hence, the circle meets the x-axis at M = −4 + 0i and
N = 6 + 0i.
We need to translate the line Γ so that M goes to the origin. Thus, we need to translate
by 4 and we need to apply the isometry T4 : (x, y) → (x + 4, y). Then, P 0 = T4 (P ) = (8, 4)
and Q0 = T4 (Q) = (9, 3). Now, we need to invert in the unit circle and need to find the
images of P 0 and Q0 . We know what Φ does:
µ
¶ µ
¶
8 4
1 1
Φ(P ) = Φ((8, 4)) =
,
=
,
80 80
10 20
µ
¶ µ
¶
9 3
1 1
0
Φ(Q ) = Φ((9, 3)) =
,
=
,
90 90
10 30
0
Note that we now have these two images on a vertical (Euclidean) line. So the distance
between the points dP (Φ(P 0 ), Φ(Q0 )) = ln(1/20) − ln(1/30) = ln(3/2). Thus, the points P
and Q are the same distance apart.
T4 (Γ)
Γ
T4(P)
4+4i
T4(Q)
5+3i
N
M
-4
-2
2
4
6
8
10
Figure 9.3: Isometries in H
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9.6
Fractional Linear Transformations
We want to be able to classify all of the isometries of the Poincaré half plane. It turns out
that the group of direct isometries is easy to describe. We will describe them and then see
why they are isometries.
A fractional linear transformation is a function of the form
T (z) =
az + b
cz + d
where a, b, c, and d are complex numbers and ad − bc 6= 0. The domain of this function is
the set of all complex numbers C together with the symbol, ∞, which will represent a point
at infinity. Extend the definition of T to include the following
T (−d/c) = lim
z→− dc
az + b
= ∞,
cz + d
if c 6= 0,
az + b
a
=
if c 6= 0,
z→∞ cz + d
c
az + b
= ∞ if c = 0.
T (∞) = lim
z→∞ cz + d
T (∞) = lim
The fractional linear transformation, T , is usually represented by a 2 × 2 matrix
·
¸
a b
γ=
c d
and write T = Tγ . The matrix representation for T is not unique, since T is also represented
by
·
¸
ka kb
kγ =
kc kd
for any scalar k 6= 0. We define two matrices to be equivalent if they represent the same
fractional linear transformation. We will write γ ≡ γ 0 .
Theorem 9.1
Tγ1 γ2 = Tγ1 (Tγ2 (z)).
From this the following theorem follows.
Theorem 9.2 The set of fractional linear transformations forms a group under composition
(matrix-multiplication).
Proof: Theorem 9.1 shows us that this set is closed under our operation. The identity
element is given by the identity matrix,
·
¸
1 0
I=
.
0 1
The fractional linear transformation associated with this is
TI (z) =
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9.6. FRACTIONAL LINEAR TRANSFORMATIONS
73
The inverse of an element is
Tγ−1 = Tγ −1 ,
since
Tγ (Tγ −1 (z)) = TI (z) = z.
We can also see that to find Tγ −1 we set w = Tγ (z) and solve for z.
az + b
cz + d
(cz + d)w = az + b
dw − b
z=
.
−cw + a
w=
That is Tγ −1 is represented by
·
¸
·
¸
1
d −b
d −b
≡
= γ −1 .
−c a
ad − bc −c a
Here we must use the condition that ad − bc 6= 0.
In mathematical circles when we have such an interplay between two objects — matrices
and fractional linear transformations — we will write γz when Tγ (z) is meant. Under this
convention we may write
·
¸
az + b
a b
.
γz =
z=
c d
cz + d
This follows the result of Theorem 9.1 in that
(γ1 γ2 )z = γ1 (γ2 z),
however in general k(γz) 6= (kγ)z. Note that
k(γz) =
k(az + b)
,
cz + d
while
(kγz) = γz =
az + b
.
cz + d
Recall the following definitions:
½·
¸
¾
a b
M2×2 (R) =
| a, b, c, d ∈ R
c d
GL2 (R) = {γ ∈ M2×2 (R) | det(γ) 6= 0}
SL2 (R) = {γ ∈ GL2 (R) | det(γ) = 1}
where R is any ring — we prefer it be the field of complex numbers, C, the field of real
numbers, R, the field of rational numbers, Q, or the ring of integers Z. GL2 (R) is called
the general linear group over R, and SL2 (R) is called the special linear group over R.
There is another group, which is not as well known. This is the projective special
linear group denoted by PSL2 (R). PSL2 (R) is obtained from GL2 (R) by identifying γ with
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kγ for any k 6= 0. The group PSL2 (C) is isomorphic to the group of fractional linear
transformations.
Remember that we wanted to classify the group of direct isometries on the upper half
plane. We want to show that any 2 × 2 matrix with real coefficients and determinant 1
represents a fractional linear transformation which is an isometry of the Poincaré upper
half plane.
Lemma 9.5 The horizontal translation by a
Ta (x, y) = (x + a, y),
can be thought of as a fractional linear transformation, represented by an element of SL2 (R).
Proof: If a ∈ R, then
Ta (x, y) = Ta (z) = z + a,
and this is represented by
z ∈ C,
·
¸
1 a
τa =
.
0 1
This is what we needed.
Lemma 9.6 The map
µ
ϕ(x, y) =
−x
y
, 2
2
2
x + y x + y2
¶
,
which is inversion in the unit circle followed by reflection through x = 0, can be thought of
as a fractional linear transformation which is represented by an element of SL2 (R).
Proof: As a function of complex numbers, the map ϕ is
ϕ(z) = ϕ(x + iy) =
This map is generated by
−x + iy
1
−(x − iy)
=− .
=
2
2
x +y
(x + iy)(x − iy)
z
·
¸
0 −1
σ=
.
1 0
Theorem 9.3 The group SL2 (R) is generated by σ and the maps τa for a ∈ R.
Proof: Note that
so
MATH 6118-090
·
¸·
¸
0 −1 1 r
στr =
1 0
0 1
·
¸
0 −1
=
1 r
·
¸·
¸
0 −1 0 −1
στs στr =
1 s
1 r
·
¸
−1
−r
=
s rs − 1
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9.7. CROSS RATIO
75
and
¸·
¸
·
0 −1 −1
−r
στt στs στr =
1 t
s rs − 1
·
¸
−s
1 − rs
=
st − 1 rst − r − t
What this means is that for any
·
γ=
¸
a b
∈ SL)2 (R)
c d
and a 6= 0, then set s = −a, solve b = 1 − rs = 1 + ra and c = st − 1 = −at − 1, giving
r=
b−1
a
and t =
−1 − c
.
a
Since det(γ) = 1, this forces d = rst − r − t. Thus, if a 6= 0, then γ can be written as a
product involving only σ and translations. If a = 0, then c 6= 0, since ad − bc = 1, and
hence
·
¸
−c −d
σγ =
,
a
b
which can be written as a suitable product. Thus SL2 (R) is generated by the translations
and σ.
Lemma 9.7 The group SL2 (R), when thought of as a group of fractional linear transformations, is a subgroup of the isometries of the Poincaré upper half plane.
Lemma 9.8 If γ ∈ GL2 (R) and detγ > 0, then γ is an isometry of the Poincaré upper half
plane.
Theorem 9.4 The image of a circle or line in C under the action of a fractional linear
transformation γ ∈ SL2 (C) is again a circle or a line.
9.7
Cross Ratio
This concept is apparently what Henri Poincaré was considering when he discovered this
particular representation of the hyperbolic plane.
S
Let a, b, c, d be elements of the extended complex numbers, C {∞}, at least three of
which are distinct. The cross ratio of a, b, c, and d is defined to be
a−c
a
(a, b; c, d) = − d .
b−c
b−d
The algebra for the element ∞ and division by zero is the same as it is for fractional linear
transformations.
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S
If we fix three distinct elements a, b, and c ∈ C {∞}, and consider the fourth element
as a variable z, then we get a fractional linear transformation:
z−b
T (z) = (z, a; b, c) = z − c .
a−b
a−c
This is the unique fractional linear transformation T with the property that
T (a) = 1,
T (b) = 0,
and T (c) = ∞.
We need to look at several examples to see why we want to use the cross ratio.
Example 9.1 Find the fractional linear transformation which sends 1 to 1, −i to 0 and
−1 to ∞.
From above we need to take: a = 1, b = −i, and c = −1. Thus, set
w = (z, 1; −i, −1)
z+i 1+i
=
/
z+1 1+1
2z + 2i
=
(1 + i)(z + 1)
In matrix notation,
·
¸
2
2i
w=
z.
1+i 1+i
Example 9.2 Find the fractional linear transformation which fixes i, sends ∞ to 3, and 0
to −1/3.
This doesn’t seem to fit our model. However, let
γ1 z = (z, i; ∞, 0)
and
γ2 w = (w, i; 3, −1/3).
So, γ1 (i) = 1, γ1 (∞) = 0, γ1 (0) = ∞, γ2 (i) = 1, γ2 (3) = 0, and γ2 (−1/3) = ∞. Therefore,
γ2−1 (1) = i, γ2−1 (0) = 3, and γ2−1 (∞) = −1/3. Now, compose these functions:
γ = γ2−1 γ1 .
Let’s check what γ does: γ(i) = i, γ(∞) = 3 and γ(0) = −1/3, as desired.
Now, set w = γ(z) and
w = γ2−1 γ1 (z)
γ2 (w) = γ1 (z)
(w, i; 3, −1/3) = (z, i; ∞, 0).
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Now, we need to solve for z:
w−3
i−3
z−∞ i−∞
/
=
/
w + 1/3 i + 1/3
z−0 i−0
(3i + 1)w − 3(3i + 1)
i
=
3(i − 3)w + i − 3
z
(−3(3i + 1)w − (3i + 1)
z=
(3i + 1)w − 3(3i + 1)
3w + 1
=
−w + 3
·
¸
3 1
=
−1 3
Then, using our identification, we will get that
·
¸ ·
¸
1 3 −1
3 −1
w=
≡
z
1 3
10 1 3
9.8
Translations
Now, we have claimed that the Poincaré upper half plane is a model for the hyperbolic
plane. We have not checked this. Let’s start with the sixth axiom:
6. Given any two points P and Q, there exists an isometry f such that f (P ) = Q.
Let P = a + bi and Q = c + di. We have many choices. We will start with an isometry
that also fixes the point at ∞. In some sense, this is a nice isometry, since it does not map
any regular point to infinity nor infinity to any regular point. Now, since f (∞) = ∞ and
f (P ) = Q, f must send the line through P and ∞ to the line through Q and ∞. This
means that the vertical line at x = a is sent to the vertical line at x = c. Thus, f (a) = c.
This now means that we have to have
(w, c + di; c, ∞) = (z, a + bi; a, ∞)
z−a
w−c
=
di
bi
d(z − a)
w=
+c
· b
¸
d bc − ad
=
z.
0
b
Since b > 0 and d > 0, then the determinant of this matrix is positive. That and the fact
that all of the entries are real means that it is an element of PSL2 (R) and is an isometry of
the Poincaré upper half plane.
We claim that this map that we have chosen is a translation. Now, recall that translations are direct isometries with no fixed points. How do we show that it has no fixed
points? A fixed point would be a point z0 so that f (z0 ) = z0 . If this is the case, then solve
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for z below:
d(z0 − a)
+ c = z0
b
ad − bc
z0 =
d−b
But, note that a, b, c, and d are all real numbers. Thus, if b 6= d then z0 is a real number
and is not in the upper half plane. Thus, this map has no fixed points in H and is a
translation. If b = d, then z0 = ∞, and again there are no solutions in the upper half plane,
so the map is a translation.
In the Poincaré upper half plane, we classify
S our translations by how many fixed points
there are on the line at infinity (that is, in R ∞.) Let
·
¸
a b
γ=
.
c d
Then γ(z) = z if
cz 2 + (d − a)z − b = 0.
Now, if c 6= 0, then this is a quadratic equation with discriminant
∆ = (d − a)2 − 4bc.
Thus, there is a fixed point in H if ∆ < 0, and no fixed points if ∆ ≥ 0. If ∆ = 0 then
there is exactly one fixed point on the line at infinity. In this case the translation is called
a parabolic translation. If ∆ > 0 the translation is called a hyperbolic translation.
9.9
Rotations
What are the rotations in the Poincaré upper half plane? What fractional linear transformations represent rotations?
A rotation will fix only one point. Let P = a + bi. We want to find the rotation that
fixes P and rotates counterclockwise through an angle of θ.
First, find the (Euclidean) line through P which makes an angle θ with the vertical line
through P . Find the perpendicular to this line, and find where it intersects the x-axis. The
circle centered at this intersection and through P is the image of the vertical line under the
rotation. Let this circle intersect the x-axis at points M and N . Then the rotation is given
by
(w, P ; N, M ) = (z, P ; a, ∞).
We want to find an easy point to rotate, then we can do this in general. It turns out
that the simplest case is to rotate about P = i.
Here let the center of the half circle be at −x, and let the (Euclidean)
radius of the circle be r. Then x =
r cos θ, r sin θ = 1, M = −r − x, and
θ
P=i
N = r − x. So we have to solve
(w, i; r − x, −r − x) = (z, i; 0, ∞).
r
1
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After quite a bit of algebraic manipulation, we get
·
¸
sin 2θ
cos 2θ
w = ρθ z =
z
− sin 2θ cos 2θ
For an arbitrary point P = a + bi we need to apply a translation that sends P to i and
then apply the rotation, and then translate back. The translation from P = a + bi to 0 + i
is
·
¸
1 −a
γ=
.
0 b
The inverse translation is
·
γ
−1
¸
b a
=
.
0 1
Thus, the rotation about P is
·
γ
9.10
−1
¸·
¸·
¸
b a
cos 2θ
sin 2θ 1 −a
ρθ γ =
0 1 − sin 2θ cos 2θ 0 b
·
¸
b cos 2θ − a sin 2θ (a2 + b2 ) sin 2θ
=
− sin 2θ
a sin 2θ + b cos 2θ
Reflections
Not all isometries are direct isometries. We have not yet described all of the orientationreversing isometries of the Poincaré upper half plane. We did see that the reflection through
the imaginary axis is given by
R0 (x, y) = (−x, y),
which is expressed in complex coordinates as
R0 (z) = −z.
Note that in terms of a matrix representation, we can represent R0 (z) by
·
¸
−1 0
R0 (z) = µz =
z
0 1
Now, to reflect through the line ` in H , first use the appropriate isometry, γ1 to move
the line ` to the imaginary axis, then reflect and move the imaginary axis back to `:
γ1−1 µγ1 z = γ1−1 µγ1 z.
Note that µ2 = 1 and that µγµ ∈ SL2 (R) for all γ ∈ SL2 (R), since detµ = −1. Therefore,
γ1−1 µγ1 z = γ1−1 (µγ1 µ)µz = γ2 µz = γ2 (−z),
where γ2 ∈ SL2 (R). Thus, every reflection can be written in the form γ(−z) for some
γ ∈ SL2 (R).
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Theorem 9.5 Every isometry f of H which is not direct can be written in the form
f (z) = γ(−z)
for some γ ∈ SL2 (R). Furthermore, if
·
¸
a b
γ=
c d
then f (z) is a reflection if and only if a = d.
9.11
Distance and Lengths
We want a formula for the distance between two points or the length of any line segment.
We have this for two points on the same vertical line. If P = a + bi and Q = a + ci, then
¯Z c ¯
¯
dy ¯¯
|P Q| = ¯¯
¯
b y
= | ln(c/b)|
Now, maybe P and Q don’t lie on a vertical line segment. Then there is a half circle with
center on the x-axis which goes through both P and Q. Let this half circle have endpoints
M and N . Since isometries preserve distance, we will look at the image of σ which sends
P to i and P Q to a vertical line. This is the transformation that sends P to i, M to 0 and
N to ∞. Since the image of Q will lie on this line, Q is sent to some point 0 + ci for some
c. Then
|P Q| = | ln(c/1)| = | ln(c)|.
Note that
(σz, i; 0, ∞) = (z, P ; M, N )
σz
and in particular, since σ(Q) = ci and (σz, i; 0, ∞) =
, we get
i
c = (Q, P ; M, N ),
so
|P Q| = | ln(Q, P ; M, N )|.
9.12
The Hyperbolic Axioms
We have not checked yet that the Poincaré upper half plane really meets all of the axioms
for a hyperbolic geometry. We need to check that all of the axioms are valid.
Axiom 1: We can draw a unique line segment between any two points.
Axiom 2: A line segment can be continued indefinitely.
We checked earlier that Axiom 2 is satisfied. Since there exists a half circle or vertical
line through any two points in the plane.
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Axiom 3: A circle of any radius and any center can be drawn.
This follows from the definition. Once we know how to measure distance, we may create
circles.
Axiom 4: Any two right angles are congruent.
Our isometries preserve Euclidean angle measurement, so define the angle measure in H
to be the same as the Euclidean angle measure. Then any two right angles are congruent.
Axiom 6: Given any two points P and Q, there exists an isometry f such that f (P ) = Q.
Axiom 7: Given a point P and any two points Q and R such that |P Q| = |P R|, there is
an isometry which fixes P and sends Q to R.
Axiom 8: Given any line `, there exists a map which fixes every point in ` and leaves no
other point fixed.
Those we established in our last 4 sections.
Axiom 5: Given any line ` and any point P 6∈ `, there exist two distinct lines `1 and `2
through P which do not intersect `.
This follows easily using non-vertical Poincaré lines.
9.13
The Area of Triangles
We have shown previously that the area of an asymptotic triangle is finite. It can be shown
that all trebly asymptotic triangles are congruent. This means that the area of all trebly
asymptotic triangles is the same. What is this common value in the Poincaré upper half
plane?
First, let’s compute the area of a doubly asymptotic triangle. We want to compute the
area of the doubly asymptotic triangle with vertices at P = ei(π−θ) in H , and vertices at
infinity of 1 and ∞. The angle at P for this doubly asymptotic triangle has measure θ.
Consider Figure 9.4.
The area element for the
Poincaré upper half plane model
is derived by taking a small
(Euclidean) rectangle with sides
oriented horizontally and vertiθ
cally. The sides approximate
P
hyperbolic segments, since the
rectangle is very small. The
area would then be a product of
the height and width (measured
with the hyperbolic arclength element). The vertical sides of the
θ
rectangle have Euclidean length
1
0
- cos θ
∆y, and since y is essentially unchanged, the hyperbolic length
Figure 9.4: Doubly Asymptotic Triangle
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∆y
∆x
. The horizontal sides have Euclidean length ∆x and hence hyperbolic length
.
y
y
dxdy
This means that the area element is given by
.
y2
is
Lemma 9.9 The area of a doubly asymptotic triangle P ΩΘ with points Ω and Θ at infinity
and with angle ΩP Θ = P has area
|4P ΩΘ| = π − P,
where P is measured in radians.
Proof: Let the angle at P have measure θ. Then 4P ΩΘ is similar to the triangle in
Figure 9.4 and is hence congruent to it. Thus, they have the same area. The area of the
triangle in Figure 9.4 is given by
Z 1
Z ∞
1
dxdy
A(θ) =
√
2
1−x2 y
− cos θ
Z 1
dx
√
=
1 − x2
− cos θ
= arccos(−x)|1− cos θ = π − θ
Corollary 3 The area of a trebly asymptotic triangle is π.
P
Ω
Θ
Σ
Figure 9.5: Trebly Asymptotic Triangle
Proof: : Let 4ΩΘΣ be a trebly asymptotic triangle, and let P be a point in the interior.
Then
|4ΩΘΣ| = |4P ΩΣ| + |4P ΘΣ| + |4P ΩΘ|
= (π − ∠ΩP Σ) + (π − ∠ΘP Σ) + (π − ∠ΩP Θ)
= 3π − 2π = π
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83
Corollary 4 Let 4ABC be a triangle in H with angle measures A, B, and C. Then the
area of 4ABC is
|4ABC| = π − (A + B + C),
where the angles are measured in radians.
In the figure below, the figure on the left is just an abstract picture from the hyperbolic
plane. The figure on the right comes from the Poincaré model, H .
Σ
A
A
B
C
C
Θ
B
Ω
Ω
Θ
Σ
Proof: Construct the triangle 4ABC and continue the sides as rays AB, BC, and CA.
Let these approach the ideal points Ω, Θ, and Σ, respectively. Now, construct the common
parallels ΩΘ, ΘΣ, and ΣΩ. These form a trebly asymptotic triangle whose area is π. Thus,
|4ABC| = π − |4AΣΩ| − |4BΩΘ| − |4CΘΣ|
= π − (π − (π − A)) − (π − (π − B)) − (π − (π − C))
= π − (A + B + C).
9.14
The Poincaré Disk Model
Consider the fractional linear transformation in matrix form
·
¸
1 −i
φ=
−i 1
or
w=
z−i
.
1 − iz
This map sends 0 to −i, 1 to 1, and ∞ to i. This map sends the upper half plane to the
interior of the unit disk. The image of H under this map is the Poincaré disk model, D.
Under this map lines and circles perpendicular to the real line are sent to circles which
are perpendicular to the boundary of D. Thus, hyperbolic lines in the Poincaré disk model
are the portions of Euclidean circles in D which are perpendicular to the boundary of D.
There are several ways to deal with points in this model. We can express points in terms
of polar coordinates:
D = {reiθ | 0 ≤ r < 1}.
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We can show that the arclength segment is
√
2 dr2 + r2 dθ2
ds =
.
1 − r2
The group of proper isometries in D has a description similar to the description on H .
It is the group
½
·
¸¾
a b
Γ = γ ∈ SL2 (C) | γ =
b a
All improper isometries of D can be written in the form γ(−z) where γ ∈ Γ.
Lemma 9.10 If dp (O, B) = x, then
d(O, B) =
ex − 1
.
ex + 1
Proof: If Ω and Λ are the ends of the diameter through OB then
x = log(O, B; Ω, Λ)
OΩ · BΛ
ex =
OΛ · BΩ
BΛ
1 + OB
=
=
BΩ
1 − OB
ex + 1
OB = x
e −1
which is what was to be proven.
9.15
Angle of Parallelism
Let Π(d) denote the radian measure of the angle of parallelism corresponding to the hyperbolic distance d. We can define the standard trigonometric functions, not as before—using
right triangles—but in a standard way. Define
sin x =
cos x =
∞
X
(−1)n
n=0
∞
X
(−1)n
n=0
x2n+1
(2n + 1)!
(9.1)
x2n
(2n)!
(9.2)
sin x
.
(9.3)
cos x
In this way we have avoided the problem of the lack of similarity in triangles, the premise
upon which all of real Euclidean trigonometry is based. What we have done is to define these
functions analytically, in terms of a power series expansion. These functions are defined for
all real numbers x and satisfy the usual properties of the trigonometric functions.
tan x =
Theorem 9.6 (Bolyai-Lobachevsky Theorem) In the Poincaré model of hyperbolic geometry the angle of parallelism satisfies the equation
µ
¶
Π(d)
−d
e = tan
.
2
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Proof: By the definition of the angle of parallelism, d = dp (P, Q) for some point P to its
foot Q in some p-line `. Now, Π(d) is half of the radian measure of the fan angle at P , or
is the radian measure of ∠QP Ω, where P Ω is the limiting parallel ray to ` through P .
We may choose ` to be a diameter of the unit disk and Q = O, the center of the disk,
so that P lies on a diameter of the disk perpendicular to `.
The limiting parallel ray through P is the arc of a circle δ so that
(a) δ is orthogonal to Γ,
(b) ` is tangent to δ at Ω.
P
α
Ω
α
Λ
Q
α
P
α
Q
Ω
Λ
Figure 9.6: Angle of Parallelism: left in D, right in H
The tangent line to δ at P must meet ` at a point R inside the disk. Now ∠QP Ω =
∠QΩP = β radians. Let us denote Π(d) = α. Then in 4P QΩ, α + 2β = π2 or
β=
Now, d(P, Q) = r tan β = r tan
¡π
ed =
Using the identity tan( π4 − α2 ) =
4
−
α
2
π α
− .
4
2
¢
. Applying Lemma 9.10 we have
r + d(P, Q)
1 + tan β
=
.
r − d(P, Q)
1 − tan β
1 − tan α/2
it follows that
1 + tan α/2
ed =
1
.
tan α/2
Simplifying this it becomes
µ
e
−d
= tan
Π(d)
2
¶
.
Also, we can write this as Π(d) = 2 arctan(e−d ).
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